Differentiability implies continuity in several variables. Is this proof correct?

Question

Let $f: R^m \to R^n$ be differentiable at $a$, then $f$ is continuous at $a$

The hint says that if $T$ is a linear map, then there exists $M > 0$ such that $||T(h)|| \leq M||h||$ for all $h \in R^m$

This is what I've done. $$ \lim_{h \to 0} ||f(a+h) - f(a) - Df(a)(h)|| = \lim_{h \to 0}||h|| * \lim_{h\to 0} \frac{||f(a+h) - f(a) - Df(a)(h)||}{||h||} = 0$$

So there exists $\delta_1$ such that if $$ ||h|| < \delta_1 \to ||f(a+h) - f(a) - Df(a)(h) || < \epsilon / 2$$

Using $\bigg| ||x|| - ||y|| \bigg| \leq ||x-y||$ with $ f(a+h) - f(a)$ as $x$ and $Df(a)(h)$ as $y$ we have

$$\bigg| | |f(a+h) - f(a)|| - ||Df(a)(h)| | \bigg| \leq ||f(a+h) - f(a) - Df(a)(h) || < \epsilon / 2 $$

Thus $$-\epsilon/2 + ||Df(a)(h)| |< | |f(a+h) - f(a)|| < \epsilon/2 + ||Df(a)(h)| |$$

This is where I'm not sure if the proof is correct

$$ | |f(a+h) - f(a)|| < \epsilon/2 + ||Df(a)(h)| | \leq \epsilon/2 + M||h||$$ So if $||h|| < \frac{\epsilon}{2M}$ we get $| |f(a+h) - f(a)|| < \epsilon/2 + \epsilon/2 = \epsilon $. All we have to do is choose $\delta = min(\delta_1, \frac{\epsilon}{2M})$

But if that's the case, the hint would be unnecessary because $Df(a)(h)$ , being a linear transformation, is continuous and $Df(a)(0) = 0$ so there is a $\delta_2$ that guarantees the inequality $ ||Df(a)(h)|| < \epsilon /2$. so we choose $\delta = min(\delta_1, \delta_2)$ and we're done.

I also found this: How to show differentiability implies continuity for functions between Euclidean spaces Copper.hat uses another approach, He writes "The limit($f(a+h) - f(a) - \lambda(h) \to 0$ as $h \to 0$) shows that for any $\epsilon > 0$ there exists a $\delta > 0$ so that if $||h||< \delta$ then $\|f(a+h) - f(a) - \lambda(h)\| \leq \epsilon \|h\|" $

I don't understand why this inequality is correct, $h$ depends on $\delta$ and $\delta$ depends on $\epsilon$, or in this case $\epsilon ||h||$ so it seems circular to me. Any help is appreciated.

Answer 1

I guess you can say the hint is unnecessary. It's just there to remind you that linear transformations (on finite dimensional vector spaces) are continuous/bounded at $0$. Perhaps it would have been easier if the hint said "If $T$ is a linear map, then it is continuous at $0$".

To expand on that, when you write $\delta = min(\delta_1, \frac{\epsilon}{2M})$, in particular the quantity $\frac{\epsilon}{2M}$, this is really the value $\delta_2$ that you mention later (appropriately scaled), except that instead of being an abstract symbol ($\delta_2$), it is rather a formula in terms of $\epsilon, M$.

As for the other issue about the inequality from the link being correct, I think there's some confusion around the meaning of "depends on" in "$h$ depends on $\epsilon$", at the least it is a different sense than the meaning in "$\delta$ depends on $\epsilon$". I'm not sure, perhaps thinking about a particular case of $\epsilon$, eg, $\epsilon=1$, may help (there exists $\delta > 0$ such that $\|h\| < \delta$ implies $\|f(a+h)-f(a)-\lambda(h)\| < \|h\|$)